# does any solid pbcl2 form when 3.5mgNacl is dissolved in 0.250L of 0.12M of lead nitrate?

## 2NaCl aq +Pn(NO32)aq >>PbCl2 (s)+2Na +2 NO3

before the lead chloride forms, lead and chloride ions are in solution. for that case...

ok, we have 250g water, so solubility is 3 grams.
we had grams of lead: .250*.12*207= 6.21, lets look at the chloride.
grams chloride: .035*35/58=.02grams Cl

mass of chloride ion in 3 grams (max allowed) of PbCL2, 3*45*2/278=.755
but we don't have that much CL ion supplied in solution, so no, no sold PbCl2 forms

## To determine if any solid PbCl2 forms when NaCl is dissolved in lead nitrate, we need to check if a precipitation reaction occurs.

First, we need to write the balanced chemical equation for the reaction between NaCl and lead nitrate:

Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3

From the equation, we can see that two moles of NaCl react with one mole of Pb(NO3)2 to produce one mole of PbCl2.

Given:
Mass of NaCl = 3.5 mg = 0.0035 g
Volume of lead nitrate solution = 0.250 L
Concentration of lead nitrate solution = 0.12 M

Next, we need to calculate the moles of NaCl and the moles of Pb(NO3)2 present in the reaction.

Moles of NaCl:
moles = mass / molar mass
moles = 0.0035 g / (22.99 g/mol + 35.45 g/mol)
moles = 0.0035 g / 58.44 g/mol
moles ≈ 0.000060 mol

Moles of Pb(NO3)2:
moles = concentration × volume
moles = 0.12 mol/L × 0.250 L
moles = 0.030 mol

Comparing the mole ratios from the balanced equation, we see that the ratio between NaCl and Pb(NO3)2 is 2:1. Thus, we need twice as many moles of Pb(NO3)2 compared to NaCl.

Since we have 0.000060 mol of NaCl, we need 2 × 0.000060 mol = 0.00012 mol of Pb(NO3)2 to fully react with the NaCl.

From the calculation, we have 0.030 mol of Pb(NO3)2, which is much more than the 0.00012 mol required to react with the NaCl.

Therefore, all the Pb(NO3)2 will react, and no solid PbCl2 will form.

## To determine whether a solid PbCl2 forms when NaCl is dissolved in lead nitrate, we can use stoichiometry and solubility rules.

First, let's write down the balanced chemical equation for the reaction between lead nitrate (Pb(NO3)2) and sodium chloride (NaCl):

Pb(NO3)2 + 2 NaCl → PbCl2 + 2 NaNO3

From the equation, we see that 1 mole of lead nitrate reacts with 2 moles of sodium chloride to form 1 mole of solid PbCl2.

Next, let's calculate the number of moles of NaCl in the solution:

Given mass of NaCl = 3.5 mg
Molar mass of NaCl = 58.44 g/mol

Number of moles of NaCl = (3.5 mg / 1000) / 58.44 g/mol

Now, let's calculate the number of moles of Pb(NO3)2 in the solution:

Volume of solution = 0.250 L
Concentration of Pb(NO3)2 = 0.12 M

Number of moles of Pb(NO3)2 = Concentration × Volume

Finally, compare the number of moles of NaCl with the stoichiometry of the reaction. If the number of moles of NaCl is greater than twice the number of moles of Pb(NO3)2, then not all of the NaCl will react and solid PbCl2 will form. Otherwise, all the NaCl will dissolve, and no solid PbCl2 will form.

Remember to convert between units and use appropriate sig figs throughout the calculations.

Please provide the values for volume or concentration of lead nitrate to proceed with the complete calculation.